AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be sequences of real numbers for which 
and
for all positive integers 
. Prove that is eventually increasing (that is, there exists a positive integer 
for which 
for all 
).
+1 w
and 
be two concentric circles. Let 
and 
be any two equilateral triangles inscribed in and respectively. If 
and 
are any two points on and respectively, show that ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
+1 w
, 
, 
, 
and 
. Extend 
to meet the circumcircle of triangle 
at 
. Prove that 
. people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by ?
with 
. Points 
and are on 
and 
respectively such that 
. Lines 
and 
meet at 
. 
is the incenter of 
and 
is the orthocenter of 
. 
and 
are circles with diameter 
and 
, respectively, intersecting each other at points 
and 
. Prove that and lie on 
.
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers such that all unit squares on the board can be made into all black after a finite number of moves.
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
such that the inequality 
holds for all positive integers 
.
be the assertion.
so 
, so 
, so 
, thus we have 
for all x.
, thus the only function is 
such that the inequality holds for all positive integers .
, you get 
and 
so . Now, we try to plug 
and 
respectively (not in once), and we get something like 
and 
. From the first equation it follows 
. Since we know that 
maps natural numbers to natural numbers, and we know that that equation works for all natural numbers 
, we can use our previous observation and plug 
, so we get something like 
so 
and 
must be 
. Ok, so now we look back again to previous equations and from it follows 
and from the 
so that means 
, but since 
, we get also for 
, combining these two things we get 
. Now from 
and we know 
so 
and hence is the only possibility. It's easy to check. Hope that helps
be the assertion of and to the given functional inequality. We will prove that only 
works. It is easy to check that this function holds.
.
gives us 
which forces 
since 
. Now, 
gives us 
, which forces .
for all 
gives us 
for all .
for all .![\[ f(x) + yf(x) \le f(x) + yf(f(x)) \le x(1 + f(y)) \Rightarrow \frac{f(x)}{x} \le \frac{f(y) + 1}{y + 1} \]](http://latex.artofproblemsolving.com/d/d/a/ddaa3f4c84e4f5a907c2e6c2ea3bc61ef1d02054.png)
Let 
, we then have for all . Therefore, we are done as we conclude 
for all .
such that the inequality holds for all positive integers .
denote the assertion of the given functional equation. 
gives us that 
, so 
. Next, 
gives us that ![\[f(y)-y\geq f(1)-1\geq0\]](http://latex.artofproblemsolving.com/e/4/9/e49df266eec87242fb895a6d68c4965e8cd90197.png)
so 
for all 
. Finally, we have ![\[xf(y)+x\geq yf(f(x))+f(x)\geq yf(x)+x\]](http://latex.artofproblemsolving.com/8/0/f/80fcb74e89905425d08efd1d72ac3af83143feb7.png)
so 
for all 
, interchanging and 
gives us that 
or 
for all where 
is a fixed constant. Plugging in gives us that ![\[(c-1)(cxy+x)\leq 0\]](http://latex.artofproblemsolving.com/e/9/6/e96d00b40a745fe57b28ec73509c629335b8512d.png)
and as 
, we must have 
, implying that 
for all 
.
is. and 
give 
Then gives 
Now it is not hard to get 
So 
denote the given assertion.
We know that because the functions takes in and out natural numbers.
we have 
Now, letting 
we have 
We also have 
using we have 
But we have 
therefore the only function that satisfies the given functional equation is 
.
From symmetry, 
. So 
, i. e. 
for some 
.
is the only working solution (by putting 
, for example), so the only solution is 
. 
combining above two relations we get
but we already have 
so f(x)=x is the only solution
Prove that
,
such that 
Prove that
gives 
which gives 
Consequently, 
which then combined with 
Plugging 
whence we get 
.
yields 
implying that 
which again yields 
finishing the problem. 
.
,
for all 
combined with 
gives 
.
.
.
.
.
,
,
,
.
denote the given assertion.
.
.
.
by (*) 
.
.
.
for all 
.
.
.
.
.